Positive symmetric matrices and positive-definiteness

Question

Is a symmetric real matrix with diagonal entries strictly greater than $1$ and off-diagonal entries positive but strictly less than $1$ necessarily positive-semidefinite?

Answer 1

Nope. Just playing around with my computer, I found the matrix

[$\frac{11}{10}$ $\frac{1}{100}$ $\frac{99}{100}$]

[$\frac{1}{100}$ $\frac{11}{10}$ $\frac{99}{100}$]

[$\frac{99}{100}$ $\frac{99}{100}$ $\frac{11}{10}$]

with determinant $\frac{-25179}{31250}$.

Is this perhaps a misremembering of the definition of a diagonally dominant matrix?

Answer 2

If the sum of each row is positive, then no eigenvalue can be non-negative: take any eigenvector, and pick the largest (in magnitude) coordinate. After applying the matrix, even if the signs are all perfectly against us, still $0$ will not be crossed (or reached to).

So the correct formulation is $\geq 1$ on the diagonal, and the off-diagonals in each row sum to less than $1$.

Answer 3

By Gershgorin's Theorem we can conclude that the absolute sum of all non-diagonal entries in each row must be less than 1. Then the matrix will have positive eigenvalues for diagonal entries > 1.